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| Description: Double complement under universal class. Exercise 4.10(s) of [Mendelson] p. 231. (Contributed by NM, 8-Jan-2002.) | 
| Ref | Expression | 
|---|---|
| ddif | ⊢ (V ∖ (V ∖ 𝐴)) = 𝐴 | 
| Step | Hyp | Ref | Expression | 
|---|---|---|---|
| 1 | vex 3484 | . . . . 5 ⊢ 𝑥 ∈ V | |
| 2 | eldif 3961 | . . . . 5 ⊢ (𝑥 ∈ (V ∖ 𝐴) ↔ (𝑥 ∈ V ∧ ¬ 𝑥 ∈ 𝐴)) | |
| 3 | 1, 2 | mpbiran 709 | . . . 4 ⊢ (𝑥 ∈ (V ∖ 𝐴) ↔ ¬ 𝑥 ∈ 𝐴) | 
| 4 | 3 | con2bii 357 | . . 3 ⊢ (𝑥 ∈ 𝐴 ↔ ¬ 𝑥 ∈ (V ∖ 𝐴)) | 
| 5 | 1 | biantrur 530 | . . 3 ⊢ (¬ 𝑥 ∈ (V ∖ 𝐴) ↔ (𝑥 ∈ V ∧ ¬ 𝑥 ∈ (V ∖ 𝐴))) | 
| 6 | 4, 5 | bitr2i 276 | . 2 ⊢ ((𝑥 ∈ V ∧ ¬ 𝑥 ∈ (V ∖ 𝐴)) ↔ 𝑥 ∈ 𝐴) | 
| 7 | 6 | difeqri 4128 | 1 ⊢ (V ∖ (V ∖ 𝐴)) = 𝐴 | 
| Colors of variables: wff setvar class | 
| Syntax hints: ¬ wn 3 ∧ wa 395 = wceq 1540 ∈ wcel 2108 Vcvv 3480 ∖ cdif 3948 | 
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1795 ax-4 1809 ax-5 1910 ax-6 1967 ax-7 2007 ax-8 2110 ax-9 2118 ax-ext 2708 | 
| This theorem depends on definitions: df-bi 207 df-an 396 df-tru 1543 df-ex 1780 df-sb 2065 df-clab 2715 df-cleq 2729 df-clel 2816 df-v 3482 df-dif 3954 | 
| This theorem is referenced by: complss 4151 dfun3 4276 dfin3 4277 invdif 4279 ssindif0 4464 difdifdir 4492 | 
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